Question

The 5th term of an arithmeric sequence is 40 and 31st term is 160

a Find the sum of 17th and 19th terms?

b)Find 18th term?
() Find the sum of first 35 terms?
d) What is the sum of first 35 terms of the
arithmetic sequence with 5th tearm 43
and 31st
term 163​

Answer 1

Answer:

Part 1

fifth term=40

40=u+(n-1)d

40=u+39d.............(i)

and, 160=u+159d.....(ii)

160=u+39d

40 =u+159d

(-)    (-)  (-)

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120=-120d

d=-1

Substituting d=-1 in (i)

40=u-39

u=40+39

u=79

Therefore

17th term=79+16(-1)

=79-16

=63

and,

19th term=79+18(-1)

=79-18

=61

Thus, sum of 16th and 18th term= 61+63

=124

Part 2

18th term=19th term-d

=61-(-1) = 62

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Part 3

35th term= 79+34(-1)

=79-34=45

Thus, Sum of 35 terms= (n/2)(u+l)

=(35/2)(79+45)

=(35/2)(124)

=35*62

=2170

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Part 4

Do part 4 similarly as we did part 1...

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I will solve it then...:)

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